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Stanford Mathematics Department Math 205A Lecture Supplement #4

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Stanford Mathematics Department Math 205A Lecture Supplement #4 2 1 Borel Regular Measures We now state and prove an important regularity property of Borel regular outer measures: 1.1 Theorem. Suppose X is a topological space with the property that every closed Stanford Mathematics Department subset of X is the countable intersection of open sets (this trivially holds e.g. if X is Math 205A Lecture Supplement #4 a metric space), suppose is a Borel-regular outer measure on X, and suppose that X j1 1Vj , where (Vj ) < and Vj is open for each j 1; 2; : : :. Then D[ D 1 D Borel Regular & Radon Measures (1) (A) inf (U ) D U open;U A for each subset A X, and (2) (A) sup (C ) D C closed;C A for each -measurable subset A X. 1 Borel Regular Measures 1.2 Remark: In case X is a locally compact separable metric space (thus for each x X there is > 0 such that the closed ball B(x) fy X d (x; y) Recall that a Borel measure on a topological space X is a measure defined on 2 D 2 W Ä g is compact, and X has a countable dense subset), the condition X j1 1Vj the collection of Borel sets, and an outer measure on X is said to be a Borel- D[ D with Vj open and (Vj ) < is automatically satisfied provided (K) < for regular outer measure if all Borel sets are -measurable and if for each subset 1 1 A X there is a Borel set B A such that (B) (A). (Notice that this each compact K. Furthermore in this case we have from (2) above that D does not imply (B A) 0 unless A is -measurable and (A) < .) Also if n D 1 (A) sup (K) is a Borel measure on X, then we get a Borel regular outer measure on X D K compact;K A by defining (Y ) inf (B) where the inf is taken over all Borel sets B with D B Y , and this outer measure coincides with on all the Borel sets. (See for each -measurable subset A K with (A) < , because under the above 1 conditions on X any closed set C can be written C i1 1Ki , Ki compact. Q.1 of hw2.) D[ D Also, if is a Borel regular outer measure on X and if A X is -measurable Proof of 1.1. We assume first that (X) < . By Borel regularity of , for with (A) < , then we claim A is also Borel regular. Here A is the 1 1 any given A X we can select a Borel set B A with (B) (A), so it outer measure on X defined by D clearly suffices to check (1) in the special case when A is a Borel set. Now let ( A)(Y ) (A Y ): D \ To check this claim first observe that if E is -measurable and Y X is A f Borel sets A (1) holdsg: D W arbitrary then ( A)(Y ) (A Y ) (A Y E) (A Y E) D \ D \ \ C \ n D Trivially A contains all open sets and one readily checks that A is closed under ( A)(Y E) ( A)(Y E), hence E is also ( A)-measurable. In \ C n both countable unions and intersections, as follows: particular all Borel sets are ( A)-measurable, so it remains to prove that for each Y X there is a Borel set B Y with ( A)(B) ( A)(Y ). To If A1; A2;::: A then for any given " > 0 there are open U1; U2;::: with 2 j D Uj Aj and (Uj Aj ) 2 ". Now one easily checks j Uj ( kAk ) prove this, first use the Borel regularity of and the fact that A is measurable n Ä [ n [ j (Uj Aj ) and j Uj ( kAk ) j (Uj Aj ) so by subadditivity we have of finite measure to pick Borel sets B1 Y A and B2 A with (B1) [ n \ n \ [ n \ D ( ( )) ( N ( )) ( (Y A) and (B A) 0, and then pick a Borel set B B A with j1 1Uj kAk < " and limN j 1Uj kAk j1 1Uj 2 3 2 [ D n [ !1 \ D n \ D \ D n \ n D n ( kAk )) < ", so both kAk and kAk are in A as claimed. (B3) 0. Then Y B1 (X A) B1 (X B2) B3 (which is a Borel D [ n [ n [ \ [ \ set) and ( A)(Y ) ( A)(B1 (X B2) B3) ((A B1) (A B3)) In particular A must also contain the closed sets, because any closed set in X Ä [ n [ D \ [ \ Ä (B1) (Y A) ( A)(Y ). can be written as a countable intersection of open sets and all the open sets D \ D Math 205A Lecture Supplement 3 4 2 Radon Measures, Representation Theorem A A fA A X A Ag A are in . Thus if we let z then z is a -algebra hence D 2 W n 2 i containing all the closed sets, and hence A contains all the Borel sets. Thus A (A ( 1 Ci`)) < 2 " z n [` contains all the Borel sets and (1) is proved in case (X) < . D1 1 so for each i 1; 2; : : : there is a positive integer J (i) such that D To check (2) in case (X) < we can just apply (1) to X A: thus for i 1 n (A ( j J (i)Cij )) < 2 " each i 1; 2; : : : there is an open set Ui X A with (X) (A) 1=i n [j jÄ D n C D (X A) 1=i > (Ui ) (X) (Ci ), where Ci X Ui X (X A) A. Since A ( i1 1( j J (i)Cij )) i1 1(A ( j J (i)Cij ) (by De Morgan), this n C D D n n n D n \ D [j jÄ D[ D n [j jÄ Thus Ci A is closed and (Ci ) > (A) 1=i. implies (A C ) < "; In case (X) , to prove (1) we first take a Borel set B with (B) (A) n D 1 D and then apply the above result for finite measures to the Borel regular measure where C i1 1( j J (i)Cij ) is a closed subset of A. D\ D [j jÄ Vj , j 1; 2; : : :, thus obtaining, for given " > 0, open Uj B with Now define gi j J (i)Cij R by setting gi (x) (j 1)=i on Cij , j D j j W[j jÄ ! Á j j Ä ( Vj )(Uj B) < "2 and hence (Uj Vj B) "2 and by subadditivity n \ n Ä J (i). Then gi is clearly continuous and its restriction to C is continuous for this gives each i; furthermore by construction 0 f (x) gi (x) 1=i for each x C Ä Ä 2 ( (Uj Vj ) B) < ": and each i 1; 2; : : :, so that gi C converges uniformly to f C on C and hence [ \ n D j j f C is continuous. Since (Uj Vj ) is an open set containing B (hence A), this is the required j [ \ result. Similarly to check (2) in case (X) , we apply (2) for the finite measure D 1 2 Radon Measures, Representation Theorem case to the Borel regular measure Vj , giving closed Cj X such that j Cj A and ( Vj )(A Cj ) < "2 for each j 1; 2; : : :, and hence (A Vj In this section we work mainly in locally compact Hausdorff spaces, and for j n D \ n k1 1Ck ) < "2 , so by subadditivity and the fact that Vj X we have (A the reader’s convenience we recall some basic definitions and preliminary topo- [ D [ D N n ( k1 1Ck )) < " and hence (A) ( k1 1Ck ) " limN ( k 1Ck ) ". logical results for such spaces. [ D N Ä [ D C D !1 [ D C Since k 1Ck is closed for each N this is the required result. [ D Recall that a topological space is said to be Hausdorff if it has the property that for every pair of distinct points x; y X there are open sets U ; V with x U , 2 2 Using the above lemma we now prove Lusin’s Theorem: y V and U V Æ. In such a space all compact sets are automatically closed, 2 \ D the proof of which is as follows: observe that if x K then for each y K we 1.3 Theorem (Lusin’s Theorem.) Let be a Borel regular outer measure on a … 2 can (by definition of Hausdorff space) pick open Uy ; Vy with x Uy , y Vy topological space X having the property that every closed set can be expressed as the 2 2 and Uy Vy Æ. By compactness of K there is a finite set y1; : : : ; yN K with countable intersection of open sets (e.g. X is a metric space), let A be -measurable N\ D N 2 K j 1Vyj . But then j 1Uyj is an open set containing x which is disjoint with (A) < , and let f A R be -measurable. Then for each " > 0 there is [ D \ D 1 W ! from j Vyj and hence disjoint from K, so that K is closed as claimed. In fact ( ) [ a closed set C X with C A, A C < ", and f C continuous. we proved a bit more: that for each x K there are disjoint open sets U ; V n j … with x U and K V . Then if L is another compact set disjoint from K we 2 Proof: For each i 1; 2; : : : and j 0; 1; 2; : : : let can repeat this for each x L thus obtaining disjoint open U ; V with x U D D ˙ ˙ x x x 1 2 2 A f ((j 1)=i; j=i]; and K Vx, and then compactness of L implies x1; : : : ; xM L such that ij 9 2 D L M U M U M V A j 1 xj and then j 1 xj and j 1 xj are disjoint open sets containing so that Aij ; j 1; 2; : : :, are p.w.d.
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